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Thread: Showing existence of an operator

  1. #1
    Senior Member Dinkydoe's Avatar
    Dec 2009

    Showing existence of an operator

    I've been given the following question:

    Let (\mathcal{H},(,)) be a real Hilbert space with an orthonormal basis  \left\{e_n\right\}_{n=1}^{\infty} . Let F:[0,1]\to \mathcal{H} be continuous. Show that there exists a positive self-adjoint operator T\in B(\mathcal{H}) such that (Tx,y) = \int_0^1(F(t),x)(F(t),y)dt for all x,y\in \mathcal{H}.

    Show moreover that T is compact.

    (you may use that \int_0^1\sum_{n=N+1}^{\infty}|(F(t),e_n)|^2dt\to 0 as N\to\infty)

    I don't quite understand what must be shown here exactly. Must I find such a operator T explicitly and then show it has the desired property. If so, how? I don't quite see where the hint is useful, I guess it's meant for showing compactness of T.

    Any hint or push in the right direction is very much appreciated.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    [This is not a solution, or even a hint, but it might give you a push in the right direction.]

    To get an idea of what is going on here, start by looking at the very simplest situation, when the function F is constant, say F(t) = z\in\mathcal{H}. Then (Tx,y) = (z,x)(z,y), so Tx = (z,x)z. This is a rank-one operator (you may perhaps have seen the notation T = z\otimes z for such an operator).

    Next, take a slightly more complicated situation, where F(t) = z_j for (j-1)/n\leqslant z_j<j/n\;(1\leqslant j\leqslant n). (Admittedly, that function F is not continuous, but the idea is to use it as a step function giving a Riemann approximation to a continuous function.) Then (Tx,y) = \frac1n\sum_{j=1}^n(z_j,x)(z_j,y). So Tx = \frac1n\sum_{j=1}^n(z_j,x)z_j, or T= \frac1n\sum_{j=1}^nz_j\otimes z_j. This is again a finite-rank operator.

    Now use the Riemann approximation idea. Given a continuous function F:[0,1]\to\mathcal{H}, let z_j = F(j/n)\;(1\leqslant j\leqslant n), and try to show that as n\to\infty the operators \frac1n\sum_{j=1}^nz_j\otimes z_j defined as in the previous paragraph converge in norm to T. Then T will be a norm limit of finite-rank operators, hence compact.
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